Chapter 2



Chapter 11. Rotation of Rigid Bodies

Uniform Angular Acceleration

11-1. A cable is wrapped around a drum 80 cm in diameter. How many revolutions of this drum will cause an object attached to the cable to move a linear distance of 2 m?

What is the angular displacement? [ R = 0.04 m, s = 2 m, ( = ? ]

[pic] = 5 rad [pic]; ( = 0.796 rev

11-2. A bicycle wheel is 26 in. in diameter. If the wheel makes 60 revolutions, what linear distance will it travel? [ D = 26 in.; R = 13 in. = 1.083 ft ]

[pic]; s = (R = (377 rad)(1.083 ft); s = 408 ft

11-3. A point on the edge of a large wheel of radius 3 m moves through an angle of 370. Find the length of the arc described by the point.

[pic]; s = (R = (0.646 rad)(3 m); s = 1.94 m

11-4. A person sitting on the edge of a 6-ft diameter platform moves a linear distance of 2 ft. Express the angular displacement in radians, degrees, and revolutions. [R = 3 ft ]

[pic]; ( = 0.667 rad ; [pic]; ( = 0.106 rad ;

[pic]; ( = 38.20

11-5. An electric motor turns at 600 rpm. What is the angular velocity? What is the angular displacement after 6 s?

[pic]; ω = 62.8 rad/s

[pic]; ( = 377 rad

11-6. A rotating pulley completes 12 rev in 4 s. Determine the average angular velocity in rev/s, rpm, and in rad/s?

[pic]; f = 3.00 rev/s ; [pic]; f = 28.6 rpm

[pic]; ω = 18.8 rad/s

11-7. A bucket is hung from a rope which is wrapped many times around a circular drum of radius 60 cm. The bucket starts from rest and is lifted to a height of

20 m in 5 s. (a) How many revolutions were made by the drum.

(b) What was the average angular speed of the rotating drum?

[pic] = 33.3 rad [pic]; ( = 5.31 rev

[pic]; v = 4.00 m/s [pic]; ω = 6.67 rad/s

11-8. A wheel of radius 15.0 cm starts from rest and makes 2.00 rev in 3.00 s. (a) What is its average angular velocity in rad/s? (b) What is the final linear velocity of a point on the rim of the wheel? [pic][pic]; ω = 4.19 rad/s

11-8. (Cont.) The final angular speed is twice the average since ωo = 0: ωf = 8.38 rad/s

vf = ωfR = (8.38 rad/s)(0.15 m); vf = 1.26 m/s

11-9. A cylindrical piece of stock 6 in. in diameter rotates in a lathe at 800 rpm. What is the linear velocity at the surface of the cylinder? [ R = D/2 = 3 in. = 0.250 ft ]

[pic]; v = 20.9 ft/s

11-10. The proper tangential velocity for machining steel stock is about 70 cm/s. At what rpm should a steel cylinder 8 cm in diameter be turned in a lathe? R = (0.08 m/2) = 0.04 m

[pic]; [pic];

[pic]; f = 0.279 rev/s(60 s/min); f = 16.7 rpm

11-11. For the wheel in Problem 11-8, what is the angular acceleration? What is the linear acceleration of a point on the edge of the wheel? [ ωf = 8.38 rad/s; ωo = 0, t = 3 s ]

[pic]; α = 2.79 rad/s2

a = αR = (2.79 rad/s2)(0.15 m); a = 0.419 m/s2

11-12. A circular drum of radius 40 cm is initially rotating at 400 rpm. It is brought to a stop after making 50 revolutions. What was the angular acceleration and the stopping time?

2αθ ’ ωf2 - ωo2; ( = 50 rev(2( rad/rev) = 314 rad; f = 400 rpm = 41.9 rad/s

[pic]; α = 2.79 rad/s2

[pic]; t = 15.0 s

11-13. A belt is wrapped around the edge of a pulley that is 40 cm in diameter. The pulley rotates with a constant angular acceleration of 3.50 rad/s2. At t = 0, the rotational speed is 2 rad/s. What is the angular displacement and angular velocity of the pulley 2 s later?

θ ’ ω0t + ½αt2; R = 0.40 m/2 = 0.20 m; ωo = 0, t = 2 s, α = 3.5 rad/s.

[pic]θ ’ ω0t + ½αt2 = (2 rad/s)(2 s) + ½(3.5 rad/s2)(2 s)2; ( = 11.00 rad

ωf = ωo + αt = 2 rad/s + (3.5 rad/s2)(2 s); ωf = 9.00 rad/s

11-14. In Problem 11-13, what is the final linear speed and linear acceleration of the pulley belt as it moves around the edge of the pulley? (We must use the radius R = 0.20 m.)

v = ωf R = (9.00 rad/s)(0.200 m) ; v = 1.80 m/s

a = αR = (3.50 rad/s2)(0.200 m) ; a = 0.750 m/s2

*11-15. A wheel rotating initially at 6 rev/s undergoes a constant angular acceleration of 4 rad/s2. What is the angular velocity after 5 s? How many revolutions will the wheel make?

ωο = 2(fR = 2((6 rev/s) = 37.7 rad/s; α = 4 rad/s2; t = 5 s

ωf = ωo + αt; ωf = 37.7 rad/s + (4 rad/s2)(5 s); ωf = 57.7 rad/s

θ ’ ω0t + ½αt2; ( = (37.7 rad/s)(5 s) + ½(4 rad/s2)(5 s)2; ( = 238 rad

[pic]; ( = 38.0 rev

*11-16. A grinding disk is brought to a stop in 40 rev. If the braking acceleration was –6 rad/s2, what was the initial frequency or revolution in rev/s? [ ( = 40 rev (2() = 251 rad ]

2αθ ’ ωf2 - ωo2; [pic]; ωo = 54.9 rad/s

[pic]; f = 8.74 rev/s

*11-17. A pulley 320 mm in diameter and rotating initially at 4 rev/s receives a constant angular acceleration of 2 rad/s2. What is the linear velocity of a belt around the pulley after 8 s? What is the tangential acceleration of the belt? [ R = 0.320/2) = 0.160 m ]

ωo = 4 rev/s (2( rad/rev) = 25.1 rad/s; α = 2 rad/s2; t = 8 s

ωf = ωo + αt = 25.1 rad/s + (2 rad/s2)(8 s); ωf = 41.1 rad/s

v = ωf R = (41.1 rad/s)(0.160 m) ; v = 6.58 m/s

a = αR = (2 rad/s2)(0.160 m); a = 0.320 m/s2

*11-18. A person initially at rest 4 m from the center of a rotating platform covers distance of 100 m in 20 s. What is the angular acceleration of the platform? What is the angular velocity after 4 s?

Find linear acceleration first: [pic]

[pic]; α = 0.125 rad/s2

ωf = ωo + αt; ωf = (0.125 rad/s2)(4 s); ωf = 0.500 rad/s

Rotational Kinetic Energy; Moment of Inertia

11-19. A 2-kg mass and a 6-kg mass are connected by a light 30-cm bar. The system is then rotated horizontally at 300 rpm about an axis 10 cm from the 6-kg mass. What is the moment of inertia about this axis? What is the rotational kinetic energy?

I = ΣmR2 = (2 kg)(0.2 m)2 + (6 kg)(0.1 m)2

I = 0.140 kg m2 ω = 300 rpm = 31.4 rad/s

Ek = ½Iω2 = ½(0.140 kg m2)(31.4 rad/s)2; Ek = 69.1 J

11-20. A 1.2-kg bicycle wheel has a radius of 70 cm with spokes of negligible weight. If it starts from rest and receives an angular acceleration of 3 rad/s2, what will be its rotational kinetic energy after 4 s? (The bicycle wheel approximates a circular hoop.)

α = 3.00 rad/s2; I = mR2 = (1.2 kg)(0.70 m)2; I = 0.588 kg m2

ωf = ωo + αt = (0) + (3 rad/s2)(4 s); ωf = 12.0 rad/s

Ek = ½Iωf2 = ½(0.588 kg m2)(12.0 rad/s)2; Ek = 42.3 J

*11-21. A 16-lb grinding disk is rotating at 400 rpm. What is the radius of the disk if its kinetic energy is 54.8 ft lb? What is the moment of inertia? [ 400 rpm = 41.89 rad/s ]

m = (16 lb/32 ft/s2) = 0.500 slugs; Ek = ½Ιω2 ’ 54.8 ft lb; I = ½mR2

[pic]; I = 0.0625 slug ft2

I = ½mR2; [pic]; R = 0.500 ft or 6.00 in.

*11-22. What must be the radius of a 4-kg circular disk if it is to have the same moment of inertia as a 1-kg rod,1 m long and pivoted at its midpoint? [ ID = ½mR2; IR = (1/12)mL2 ]

[pic]; R = 0.204 m

*11-23. A wagon wheel 60 cm in diameter is mounted on a central axle where it spins at 200 rpm. The wheel can be thought of as a circular hoop of mass 2 kg, and each of 12 wooden 500-g spokes can be thought of as thin rods rotating about their ends. Calculate the moment of inertia for the entire wheel? What is the rotational kinetic energy?

mw = 2 kg, Rw = 0.30 m; ms = 0.5 kg, Ls = 0.30 m, [pic]

[pic][pic];

I = 0.360 kg m2 ω = 200 rpm = 20.94 rad/s;

[pic]; Ek = 78.9 J

*11-24. Compare the rotational kinetic energies of three objects of equal radius and mass: A circular hoop, a circular disk, and a solid sphere.

The rotational inertias are: [pic]:

For purposes of comparing, suppose we set m = 1 kg and R = 1 m

We see that: [pic]

Now Ek = ½Iω2, so at a given rotational speed, the hoop has the largest kinetic energy, followed by the disk, and then by the sphere.

Newton’s Second Law and Rotation

11-25. A rope wrapped around a 5-kg circular drum pulls with a tension of 400 N. If the radius of the drum is 20 cm and it is free to rotate about its central axis, what is the angular acceleration?

( = FR = Iα; I = ½mR2

[pic] ; α = 800 rad/s2

11-26. The flywheel of an engine has a moment inertia of 24 slug ft2. What torque is required to accelerate the wheel from rest to an angular velocity of 400 rpm in 10 s.

ωf = 400 rpm = 41.9 rad/s; ωo = 0; t = 10 s, I = 24 slug ft2

[pic];

( = Iα = (24 slug ft2)(4.19 rad/s2); ( = 1010 N m

*11-27. A 3-kg thin rod is 40 cm long and pivoted about its midpoint. What torque is required to cause it to make 20 rev while its rotational speed increases from 200 rpm to 600 rpm?

( = 20 rev(2( rad) = 126 rad; ωo = 200 rpm = 20.94 rad/s

ωf = 600 rpm = 62.8 rad/s; m = 3 kg; L = 0.40 m

2αθ ’ ωf2 - ωo2; 2α(126 rad) = (62.8 rad/s)2 – (20.94 rad/s)2; α = 13.9 rad/s2

[pic] [pic]; ( = 0.558 N m

*11-28. A large 120-kg turbine wheel has a radius of gyration of 1 m. A frictional torque of 80 N m opposes the rotation of the shaft. What torque must be applied to accelerate the wheel from rest to 300 rpm in 10 s? ωo = 0, ωf = 300 rpm = 31.4 rad/s, t = 2 s

[pic]; I = mk2 = (120 kg)(1 m)2

I = 120 kg m2; ( = Iα = (120 kg m2)(15.7 rad/s2); τ = 1885 N m

11-29. A 2-kg mass swings in a circle of radius 50-cm at the end of a light rod. What resultant torque is required to give an angular acceleration of 2.5 rad/s2.

I = mR2 = (2 kg)(0.5 m)2 = 0.5 kg m2;

τ = Iα = (0.5 kg m2)(2.5 rad/s2); τ = 1.25 N m

*11-30. A rope is wrapped several times around a cylinder of radius 0.2 m and mass 30 kg. What is the angular acceleration of the cylinder if the tension in the rope is 40 N and it turns without friction.

[pic]; α = 13.3 rad/s2

*11-31. An 8-kg grinding disk has a diameter of 60 cm and is rotating at 600 rpm. What braking force must be applied tangentially to the disk if it is to stop rotating in 5 s?

ω0 = 600 rpm = 62.8 rad/s, ωf = 0, [pic]

[pic] [pic]

F = 15.1 N

11-32. An unbalanced torque of 150 N m imparts an angular acceleration of 12 rad/s2 to the rotor of a generator. What is the moment of inertia?

τ = Iα ; [pic]; I = 12.5 kg m2

Rotational Work, Energy, and Power

11-33. A rope wrapped around a 3-kg disk 20 cm in diameter is pulled for a linear distance of 5 m with a force of 40 N. What is the linear work done by the 40-N force? What is the rotational work done on the disk? [ R = 20/2) = 10 cm or 0.10 m ]

Work = F s = (40 N)(5 m) ; Work = 200 J [pic]

Work = τ( = FR( = (40 N)(0.10 m)(50 rad); Work = 200 J

*11-34. Use the work-energy theorem to calculate the final angular velocity of the disk in Problem 11-33 if it starts from rest.

Work = ½Iωf2 – (0); I = ½mR2; Work = ½(½mR2) ωf2

[pic]; ωf = 163 rad/s

*11-35. A 1.2 kW motor acts for 8 s on a wheel having a moment of inertia of 2 kg m2. Assuming the wheel was initially at rest, what average angular speed was developed?

[pic]

Work = ½Iωf2 – (0); 9600 J = ½(2 kg m2) ωf2 ; ωf = 98.0 rad/s

*11-36. A cord is wrapped around the rim of a cylinder that has a mass of 10 kg and a radius of 30 cm. If the rope is pulled with a force of 60 N, what is the angular acceleration of the cylinder? What is the linear acceleration of the rope?

I = ½mR2 = ½(10 kg)(0.30 m)2 = 0.450 kg m2; τ = (60 N)(0.3 m) = 18 N m

[pic] α = 40 rad/s2

11-37. A 600-W motor drives a pulley with an average angular velocity of 20 rad/s. What torque is developed?

[pic] τ = 30.0 N m

11-38. The crankshaft on an automobile develops 350 lb ft of torque at 1800 rpm. What is the output horsepower? 1800 rpm = 188.5 rad/s ;

P = τω = (350 ft lb)(188.5 rad/s) = 65, 973 ft lb/s or 120 hp

Angular Momentum

11-39. A 500-g steel rod 30 cm in length is pivoted about its center and rotated at 300 rpm. What is the angular momentum? [ ω = 300 rpm = 31.4 rad/s; m = 0.5 kg ]

[pic]; I = 0.00375 kg m2

[pic]; Iω = 0.118 kg m/s2

11-40. In Problem 11-39, what average torque must be applied to stop the rotation in 2 s?

[pic]

τ = Iα = (0.00375 kg m2)(15.7 rad/s2); τ = 0.0589 N m

*11-41. A sudden torque of 400 N m is applied to the edge of a disk initially at rest. If the rotational inertia of the disk is 4 kg m2, and the torque acts for 0.02 s, what is the change in angular momentum? What is the final angular speed?

change in angular momentum = Angular impulse; τ Δt = Iωf - Iωο

τ Δt = (400 N m)(0.02 s) = 8.00 kg m/s2

τ Δt ’ Ιωf = 8.00 kg m/s2 [pic]; ωf = 2.00 rad/s

*11-42. In Fig. 11-12, a 6-kg disk A rotating clockwise at 400 rpm engages with a 3-kg disk B initially at rest. The radius of disk A is 0.4 m and the radius of disk B is 0.2 m. What is the combined angular speed after the two disks are meshed?

IA = ½(6 kg)(0.4 m)2 = 0.480 kg m2; IB = ½(3 kg)(0.2 m)2 = 0.060 kg m2

ωAo = 400 rpm = 41.9 rad/s; ωBo = 0; ωAf = ωBf = ωc;

Conservation of Momentum: Ι“ωAo + IBωBo = (IA + IB) ωc

*11-42. (Cont.) [pic]; ωc = 37.2 rad/s

*11-43. Assume that disk B in Problem 11-42 is initially rotating clockwise at 200 rpm in the same direction as disk A. What is the common angular speed as they mesh?

Choose clockwise as positive and use rpm’s for angular speed.

Ι“ωAo + IBωBo = (IA + IB) ωc

(0.48 kg m2)(400 rpm) + (0.06 kg m2)(200 rpm) = ([pic])ωc

204 kg m2 rpm = 0.54ωc; ωc = 378 rpm

*11-44. Assume the same conditions as Problem 11-42 except that the disk B is rotating counterclockwise and A is rotating clockwise. What is the combined angular velocity after the disks are meshed?

(0.48 kg m2)(400 rpm) + (0.06 kg m2)(-200 rpm) = ([pic])ωc

180 kg m2 rpm = 0.54ωc; ωc = 333 rpm

11-45. The rod connecting the two weights in Fig. 11-13 has negligible weight but is configured to allow the weights to slip outward. At the instant when the angular speed is 600 rpm, the 2-kg masses are 10 cm apart. What is the rotational speed later when the masses are 34 cm apart?

Io = (2 kg) 0.05 m)2 + (2 kg)(0.05 m)2; If = (2 kg)(0.17 m)2 + (2 kg)(0.17 m)2

I0 = 0.010 kg m2; If = 0.68 kg m2; ωo = 600 rpm ; Ioωo = Ifωf

(0.010 kg m2)(600 rpm) = (0.116 kg m2)ωf ; ωf = 51.9 rpm.

Challenge Problems

*11-46. The 6-kg circular grinding disk is rotating initially at 500 rpm. The radius of the disk is 40 cm. What is the angular acceleration of the disk if the ax exerts a tangential force of 120 N at the edge. How many revolutions will the disk make before stopping? What is work is done and what power is lost in the process? [ ω0 = 500 rpm = 52.35 rad/s ]

τ = FR = (120 N)(0.40 m) = 48 N m; I = ½mR2 = ½(6 kg)(0.4 m)2 = 0.48 kg m2

τ = Iα ; [pic]; α = -100 rad/s2

2αθ ’ ωf2 - ωo2; [pic]; ( = 13.7 rad = 2.18 rev

Work = τ( = (48 N m)(13.7 rad); Work = 658 J

[pic][pic]

[pic]P = (48 N m)(26.18 rad/s); P = 1.26 kW

*11-47 A 3-kg wheel with spokes of negligible mass is free to rotate about its center without friction. The edge of the wheel of radius 40 cm is struck suddenly with a average tangential force of 600 N lasting for 0.002 s. (a) What angular impulse is imparted to the wheel? (b) If the wheel was initially at rest, what was its angular speed at the end of the 0.002-s interval?

τ = FR = (600 N)(0.40 m) = 240 N m;

I = mR2 = (3 kg)(0.40 m)2 ; I = 0.48 kg m2

τ Δt = (240 N m)(0.002 s); τ Δt = 0.48 N m s

τ Δt = Iωf - 0; [pic] ωf = 1.00 rad/s

*11-48. Disk A has three times the rotational inertia of Disk B. Disk A is rotating initially clockwise at 200 rpm and Disk B is rotating in the opposite direction at 800 rpm. (a) If the two are meshed together, what is the common rate of rotation of the combined disks?

Assume clockwise is positive: Ι“ωAo + IB ωBo = (IA + IB) ωc ; IA = 3 IB

(3 IB)(200 rpm) + IB (-800 rpm) = (3IB + IB)ωc

-(200 rpm) IB = 4IB ωc ; [pic] ; ωc = -50.0 rpm

*11-49. What if the disks in Problem 11-48 are initially rotating in the same direction? What would be the common angular speed after meshing? (Clockwise positive)

(3 IB)(200 rpm) + IB (+800 rpm) = (3IB + IB)ωc

(1400 rpm) IB = 4IB ωc ; [pic] ; ωc = 350 rpm

*11-50. The radius of gyration of an 8-kg wheel is 50 cm. Find its moment of inertia and its kinetic energy if it is rotating at 400 rpm. ω = 400 rpm = 41.9 rad/s

I = mk2 = (8 kg)(0.5 m)2; I = 2.00 kg m2

Ek = ½Iω2 = ½(2 kg m2)(41.9 rad/s)2; Ek = 1750 J

*11-51. How much work is required to slow the wheel in Problem 11-50 to 100 rpm?

Work = change in kinetic energy

Work = ½I wf2 - ½I wo2

ωo = 41.9 rad/s; ωf = 100 rpm = 10.5 rad/s

Work = ½(2 kg m2)(10.5 rad/s)2 - ½(2 kg m2)(41.9 rad/s)2

Work = -1644 J

11-52. A wheel of radius 2 ft has a moment of inertia of 8.2 slug ft2. A constant force of 12 lb acts tangentially at the edge of the wheel that is initially at rest. What is the angular acceleration?

τ = Iα; τ = FR = (12 lb)(2 ft) = 24 lb ft;

[pic] α = 2.93 rad/s2

*11-53. In Problem 11-52, the wheel was brought to rest in 5 s. How much work was done? What horsepower was developed?

θ ’ ω0t + ½αt2 = 0 + ½(2.93 rad/s2)(5 s)2; ( = 36.6 rad

Work = τ( = (24 lb ft)(36.6 rad); Work = 878 ft lb

[pic] P = 175.7 ft lb/s or 0.319 hp

*11-54. An engine operating at 1800 rpm develops 200 hp. What is the torque developed?

ω = 1800 rpm = 188.5 rad/s; P = 200 hp = 110,000 ft lb

[pic]; τ = 584 lb ft

11-55. A constant force of 200 N acts at the edge of a wheel 36 cm in diameter causing it to make 20 rev in 5 s. What power is developed?

τ = FR = (200 N)(0.18 m); τ = 36.0 N m;

[pic]

P = τω = (36 N m)(25.13 rad/s); P = 904 W

*11-56. A 2-kg circular hoop rolls down an inclined plane from an initial height of 20 m. The kinetic energy that develops is shared between rotation and translation. What will be the speed when it reaches the bottom of the incline. [ h = 20 m, vo = ωo = 0.]

mgh = ½mv2 + ½Iω2; I = mR2; ω2 = (v2/R2)

mgh = ½mv2 + ½(mR2)(v2/R2); mgh = ½mv2 + ½mv2

v2 = gh = (9.8 m/s2)(20 m); v = 14.0 m/s

*11-57. Suppose a circular disk rolls down the same incline as in Problem 11-56. What is its speed when it reaches the bottom? (Same as Prob. 11-56 except that I = ½mR2)

mgh = ½mv2 + ½Iω2; I = ½mR2; ω2 = (v2/R2)

mgh = ½mv2 + ½(½mR2)(v2/R2); mgh = ½mv2 + ¼mv2

¾v2 = gh; [pic]; v = 16.2 m/s

Critical Thinking Questions

11-58. A circular hoop of mass 2-kg and radius 60 cm spins freely about its center connected by light central spokes. A force of 50 N acts tangent to the edge of the wheel for a time of 0.02 s. (a) What is the angular impulse? (b) What is the change in angular momentum? (c) If the hoop was initially at rest, what was the final angular speed? (d) Use the work-energy theorem to calculate the angular displacement. τ = FR; Impulse = τ Δt

τ Δt = FR Δt = (50 N)(0.6 m)(002 s) τ Δt = 0.600 N m s

Ιωf - Iωo = τ Δt; Change in momentum = 0.600 kg m2/s

I = (2 kg)(0.60 m)2 = 1.2 kg m2; Ιωf - Iωo = 0.600 kg m2/s

11-58. (Cont.) I = mR2; [pic] ωf = 0.833 rad/s

Work = τ( = FR(; FR( = ½Iωf2 – 0;

[pic]; ( = 0.00833 rad

11-59. The spin cycle on a washer slows from 900 rpm to 300 rpm in 4 s. Determine the angular acceleration. Does a force act to throw off the water from the clothes or is it the lack of a force? When operating at 900 rpm, the output power is 4 kW. What is the torque developed? If the radius of the tub is 30 cm, what is the linear speed of the clothes near the inside edge?

ωo = 90 rpm = 94.25 rad/s; ωf = 600 rpm = 62.83 rad/s; R = 0.30 m

[pic]; α = -7.86 rad/s2

P = τω ; [pic]; τ = 42.4 N m

v = ωR = (94.25 rad/s)(0.30 m); v = 27.7 m/s

11-60. A block is attached to a cord passing over a pulley through a hole in a horizontal table top as shown in Fig. 11-14. Initially the block is revolving at 4 rad/s at a distance r from the center of the hole. If the cord is pulled from below until its radius is r/4, what is the new angular velocity?

Angular momentum is conserved: Ioωo = Ifωf ; (mro2)ωo = (mrf2)ωf

[pic]; [pic] ωf = 64 rad/s

*11-61. Suppose in Fig. 11-14 the block has a mass of 2 kg and is rotating at 3 rad/s when r = 1 m. At what distance r will the tension in the cord be 25 N?

Ioωo = Ifωf ; (mro2)ωo = (mrf2)ωf ; [pic] v2 = ω2r2

[pic]

Substitute [pic] into [pic] and solve for rf :

[pic]

[pic][pic]; rf = 0.849 m or 84.9 cm

*11-62. Consider Fig. 11-15 in which m = 2 kg, M = 8 kg, R = 60 cm, and h = 6 m. Write Newton’s second law for the disk in terms of the tension in the rope, the moment of inertia of the disk, and the angular acceleration. Next write Newton’s second law for falling mass in terms the tension in the rope, the mass, and the linear acceleration. Eliminate T from these two equations. Find the linear acceleration of the 2-kg mass by recalling that v = ωR, a = αR, and I = ½mR2.

(a) τ = FR = Iα; [pic]

(b) mg – T = ma; mg – (½Ma) = ma

(c) 2mg – Ma = 2ma; 2mg = Ma + 2ma

[pic]; a = 3.27 m/s2

*11-63. Use conservation of energy to find the velocity of the 2-kg mass in Fig. 11-15 just before it strikes the floor 6 m below. Use data given in Problem 11-5.

Given: m = 2 kg, M = 8 kg, R = 60 cm, and h = 6 m, I = ½MR2

Epo of falling mass = Ekf of falling mass + Ekf of rotating disk

mgh = ½mv2 + ½Iωf2 ; I = ½MR2; ωf2 = vf2/R2

[pic]; mgh = ½mv2 + ¼Mv2

(2 kg)(9.8 m/s2)(6 m) = ½(2 kg)v2 + ¼(8 kg)v2; v2 = 39.2 m2/s2; v = 6.26 m/s

*11-64. A student stands on a platform with arms outstretched, holding weights in each hand so that the rotational inertial is 6.0 kg m2. The platform is then set into constant, frictionless rotation at 90 rpm. Now the student is able to reduce the rotational inertia to 2 kg m2 by pulling the weights in to the body. (a) What will be the new rate of rotation in the absence of external torque? (b) What is the ratio of the final kinetic energy to the initial kinetic energy? Explain increase in energy. [ ωo= 90 rpm, Io = 6 kg m2, If = 2 kg m2 ]

Conservation of momentum: Ioωo = Ifωf

[pic](6 kg m2)(90 rpm) = (2 kg m2) ωf ; ωf = 270 rpm

ωo = 90 rpm = 9.425 rad/s; ωf = 270 rpm = 28.27 rad/s

[pic]; Ek/Ef = 3.00

The final kinetic energy is three times the initial energy. The increase in energy comes for doing work on the masses to bring them closer to the body.

11-65. The large pulley shown in Fig. 11-16 can be considered as a 6-kg disk of radius 50 cm. The right mass is 4 kg and the left mass is 2 kg. Consider both rotational and translational energies and find the velocity just before the 4-kg mass strikes the floor?

Conservation of energy: The initial potential energy of the right mass

must equal the sum of the final potential and kinetic energies

including rotational Ek and translational Ek.

Three masses: m2gh = m1gh + ½m2v2 + ½m1v2 + ½Iω2

Substitute I = ½MR2 and ωf2 = v2/R2,

m2gh = m1gh + ½m2v2 + ½m1v2 + ½(½MR2)(v2/R2)

m2gh = m1gh + ½m2v2 + ½m1v2 + ¼Mv2

½m2v2 + ½m1v2 + ¼Mv2 = m2gh - m1gh

½(4 kg)v2 + ½(2 kg)v2 + ¼(6 kg)v2 = (4 kg)(9.8 m/s2)(6 m) – (2 kg)(9.8 m/s2)(6 m)

2v2 + v2 + 1.5 v2 = 117.6 m2/s2; v = 5.11 m/s

-----------------------

m1

M

R

m

h

M

R

m2

h

mg

T

T

R

h

M

m

R

h

34 cm

10 cm

2 kg

m = 3 kg

L = 0.40 m

R

6 kg

2 kg

20 cm

10 cm

0

0

t = 5 s

20 m

R

s

R

F = 400 N

0

2 kg

2 kg

2 kg

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